Simplify the following expression: $\dfrac{66t}{48t^4}$ You can assume $t \neq 0$.
Solution: $ \dfrac{66t}{48t^4} = \dfrac{66}{48} \cdot \dfrac{t}{t^4} $ To simplify $\frac{66}{48}$ , find the greatest common factor (GCD) of $66$ and $48$ $66 = 2 \cdot 3 \cdot 11$ $48 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$ $ \mbox{GCD}(66, 48) = 2 \cdot 3 = 6 $ $ \dfrac{66}{48} \cdot \dfrac{t}{t^4} = \dfrac{6 \cdot 11}{6 \cdot 8} \cdot \dfrac{t}{t^4} $ $\phantom{ \dfrac{66}{48} \cdot \dfrac{1}{4}} = \dfrac{11}{8} \cdot \dfrac{t}{t^4} $ $ \dfrac{t}{t^4} = \dfrac{t}{t \cdot t \cdot t \cdot t} = \dfrac{1}{t^3} $ $ \dfrac{11}{8} \cdot \dfrac{1}{t^3} = \dfrac{11}{8t^3} $